Given two integers x and y. In one step, we have to subtract the greatest common divisor of x and y from x and y each till x ≥ 1 and y ≥ 1. We have to find the minimum number of repetitions of this step.
Examples:
Input: x = 36, y = 16
Output: 4
Explanation: GCD of 36 and 16 is 4, so replace 36 and 16 with 32 and 12
GCD of 32 and 12 is 4, so replace 32 and 12 with 28 and 8
GCD of 28 and 8 is 4, so replace 28 and 8 with 24 and 4
GCD of 24 and 4 is 4, so replace 24 and 4 with 20 and 0
Now, as y < 1 so, process terminated. So steps repeated 4 times, so output will be 4.Input: x = 60, y = 9
Output: 3
Recursive Approach: The approach is to calculate the greatest common divisor of the two input numbers, finds the maximum possible reduction in the two numbers, and calculate the amount of reduction needed. Then recursively calls itself with updated values of the two numbers after the reduction until the two numbers become less than 1.
Steps to implement the above approach:
- Define a recursive function “solve” which takes two integers a and b as input.
- Check if either a or b is zero, and return 0 if true.
- Check if a and b are equal, and return 1 if true.
- Find the greatest common divisor of a and b using the “__gcd” function.
- Swap a and b if a is greater than b.
- Calculate the values of x and y using the formulas: x = (b-a)/g and y = a/g, where g is the greatest common divisor of a and b.
- Find the maximum value of y/i * i for all factors i of x, and call this value mx.
- Calculate the reduction in a and b after performing the operation, and call it red. The formula for red is (y – mx) * g.
- If red is greater than or equal to a, return a/g.
- If red is zero, return a/g.
- Otherwise, recursively solve the problem on the new values of a and b after performing the operation. The formula for the return value is (y – mx) + solve(a – red, b – red).
Below is the implementation of the above approach:
C++
// C++ code to solve the problem// "Common Divisor Reduction"#include <bits/stdc++.h>using namespace std;// Recursive function to// solve the problemint solve(int a, int b){ // If either a or b is zero, // we can't perform the operation if (a == 0 || b == 0) return 0; // If a and b are equal, we can // only perform the operation once if (a == b) return 1; // Find the greatest common // divisor of a and b int g = __gcd(a, b); // Swap a and b if a is // greater than b if (a > b) swap(a, b); // Calculate the values of x and y int x = (b - a) / g, y = a / g; // Find the maximum value of y/i * i // for all factors i of x, and // call this value mx int mx = 0; for (int i = 2; i * i <= x; i++) { if (x % i != 0) continue; int j = x / i; mx = max(mx, (y / i) * i); mx = max(mx, (y / j) * j); } mx = max(mx, (y / x) * x); // Calculate the reduction in a and b // after performing the operation, // and call it red int red = (y - mx) * g; // If red is greater than or equal to // a, we can't perform the // operation anymore if (red >= a) return a / g; // If red is zero, we can only perform // the operation once if (red == 0) return a / g; // Otherwise, recursively solve the // problem on the new values of a and // b after performing the operation return (y - mx) + solve(a - red, b - red);}// Driver Codeint main(){ // Example usage int x = 36, y = 16; // Function Call cout << solve(x, y); return 0;} |
Java
// JAVA code to solve the problem// "Common Divisor Reduction"import java.util.*;class GFG { public static int solve(int a, int b) { // If either a or b is zero, // we can't perform the operation if (a == 0 || b == 0) return 0; // If a and b are equal, we can // only perform the operation once if (a == b) return 1; // Find the greatest common // divisor of a and b int g = gcd(a, b); // Swap a and b if a is // greater than b if (a > b) { int temp = a; a = b; b = temp; } // Calculate the values of x and y int x = (b - a) / g, y = a / g; // Find the maximum value of y/i * i // for all factors i of x, and // call this value mx int mx = 0; for (int i = 2; i * i <= x; i++) { if (x % i != 0) continue; int j = x / i; mx = Math.max(mx, (y / i) * i); mx = Math.max(mx, (y / j) * j); } mx = Math.max(mx, (y / x) * x); // Calculate the reduction in a and b // after performing the operation, // and call it red int red = (y - mx) * g; // If red is greater than or equal to // a, we can't perform the // operation anymore if (red >= a) return a / g; // If red is zero, we can only perform // the operation once if (red == 0) return a / g; // Otherwise, recursively solve the // problem on the new values of a and // b after performing the operation return (y - mx) + solve(a - red, b - red); } public static int gcd(int a, int b) { if (b == 0) return a; return gcd(b, a % b); } public static void main(String[] args) { int x = 36, y = 16; System.out.println(solve(x, y)); }}// This code is contributed by shivamgupta310570 |
Python3
import math# Recursive function to solve the problemdef solve(a, b): # If either a or b is zero, we can't perform the operation if a == 0 or b == 0: return 0 # If a and b are equal, we can only perform the operation once if a == b: return 1 # Find the greatest common divisor of a and b g = math.gcd(a, b) # Swap a and b if a is greater than b if a > b: a, b = b, a # Calculate the values of x and y x = (b - a) // g y = a // g # Find the maximum value of y//i * i for all factors i of x, and call this value mx mx = 0 for i in range(2, int(math.sqrt(x)) + 1): if x % i != 0: continue j = x // i mx = max(mx, (y // i) * i) mx = max(mx, (y // j) * j) mx = max(mx, (y // x) * x) # Calculate the reduction in a and b after performing the operation, and call it red red = (y - mx) * g # If red is greater than or equal to a, we can't perform the operation anymore if red >= a: return a // g # If red is zero, we can only perform the operation once if red == 0: return a // g # Otherwise, recursively solve the problem on the new values of a and b after performing the operation return (y - mx) + solve(a - red, b - red)# Driver codeif __name__ == '__main__': # Example usage x = 36 y = 16 # Function call print(solve(x, y))# This code is contributed by shivamgupta310570 |
C#
using System;class GFG{ public static int Solve(int a, int b) { // If either a or b is zero, // we can't perform the operation if (a == 0 || b == 0) return 0; // If a and b are equal, we can // only perform the operation once if (a == b) return 1; // Find the greatest common // divisor of a and b int g = GCD(a, b); // Swap a and b if a is // greater than b if (a > b) { int temp = a; a = b; b = temp; } // Calculate the values of x and y int x = (b - a) / g, y = a / g; // Find the maximum value of y/i * i // for all factors i of x, and // call this value mx int mx = 0; for (int i = 2; i * i <= x; i++) { if (x % i != 0) continue; int j = x / i; mx = Math.Max(mx, (y / i) * i); mx = Math.Max(mx, (y / j) * j); } mx = Math.Max(mx, (y / x) * x); // Calculate the reduction in a and b // after performing the operation, // and call it red int red = (y - mx) * g; // If red is greater than or equal to // a, we can't perform the // operation anymore if (red >= a) return a / g; // If red is zero, we can only perform // the operation once if (red == 0) return a / g; // Otherwise, recursively solve the // problem on the new values of a and // b after performing the operation return (y - mx) + Solve(a - red, b - red); } public static int GCD(int a, int b) { if (b == 0) return a; return GCD(b, a % b); } public static void Main(string[] args) { int x = 36, y = 16; Console.WriteLine(Solve(x, y)); }} |
Javascript
// Function to calculate the gcdfunction gcd(a, b) { if (b === 0) { return a; } return gcd(b, a % b);}// Recursive function to// solve the problemfunction solve(a, b) { // If either a or b is zero, // we can't perform the operation if (a === 0 || b === 0) { return 0; } // If a and b are equal, we can // only perform the operation once if (a === b) { return 1; } // Find the greatest common // divisor of a and b let g = gcd(a, b); // Swap a and b if a is // greater than b if (a > b) { [a, b] = [b, a]; } // Calculate the values of x and y let x = (b - a) / g; let y = a / g; // Find the maximum value of y/i * i // for all factors i of x, and // call this value mx let mx = 0; for (let i = 2; i * i <= x; i++) { if (x % i !== 0) { continue; } let j = x / i; mx = Math.max(mx, (y / i) * i); mx = Math.max(mx, (y / j) * j); } mx = Math.max(mx, (y / x) * x); // Calculate the reduction in a and b // after performing the operation, // and call it red let red = (y - mx) * g; // If red is greater than or equal to // a, we can't perform the // operation anymore if (red >= a) { return a / g; } // If red is zero, we can only perform // the operation once if (red === 0) { return a / g; } // Otherwise, recursively solve the // problem on the new values of a and // b after performing the operation return (y - mx) + solve(a - red, b - red);}// Test case let x = 36, y = 16;console.log(solve(x, y)); |
Output
4
Time Complexity: O(sqrt(x)log(x))
Auxiliary Space: O(1)
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